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%%文档的题目、作者与日期
\author{王立庆（2022级数学与应用数学1班） }
\title{常微分方程期中练习解答}
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\date{2023 年 10 月 31 日}
%\date{March 9, 2021}

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\begin{document}

\maketitle

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\begin{enumerate}

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\item  %第1题
\begin{enumerate}
\item  求出平面上所有直线所满足的微分方程。
\item  求出平面上所有过原点的抛物线所满足的微分方程。
\end{enumerate}

\vspace{0.2cm}

{\color{red}解答：
\begin{enumerate}
\item  平面上的任意直线的方程为 $y=ax+b$, 其中 $a,b$ 为任意常数。求导可得 $y'=a$, $y''=0$. 因此所求微分方程为 $y''=0$. 

\item  平面上所有过原点的抛物线的方程为 $y=ax^2+bx$, 其中 $a,b$ 为任意常数。求导可得
\begin{eqnarray*}
y' &=& 2ax+b, \\
y'' &=& 2a.
\end{eqnarray*}
三个方程联立，消去任意常数 $a,b$, 得到所求微分方程为
\begin{eqnarray*}
x^2y'' -2xy'+2y=0.
\end{eqnarray*}

\end{enumerate}

}

\vspace{0.2cm}

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\item  %第2题
作出微分方程 $\frac{dy}{dx}=xy-1$ 在第一象限的线素场的草图。取下述格点。
\begin{table}[ht!]\centering
\begin{tabular}{|M{1cm}|M{1cm}|M{1cm}|M{1cm}|}\hline
$dy/dx$ & $x=0$ & $x=1$ & $x=2$ \\ \hline
$y=0$ &&& \\ \hline 
$y=1$ &&& \\ \hline 
$y=2$ &&& \\ \hline 
\end{tabular}
\end{table}

\vspace{0.2cm}

{\color{red}解答：先计算在所给格点处的 $f(x,y)$ 的值，然后以此为斜率画出短折线。
\begin{table}[ht!]\centering
\begin{tabular}{|M{1cm}|M{1cm}|M{1cm}|M{1cm}|}\hline
$dy/dx$ & $x=0$ & $x=1$ & $x=2$ \\ \hline
$y=0$ & $-1$ & $-1$ & $-1$  \\ \hline 
$y=1$ & $-1$ & $0$ & $1$ \\ \hline 
$y=2$ & $-1$ & $1$ & $3$ \\ \hline 
\end{tabular}
\end{table}

\begin{center}
\includegraphics [height=5cm, width=8cm]{ode-midterm-problem-2.png}
\end{center}

}

\vspace{0.2cm}

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\item  %第3题
判断下述方程是否为恰当方程，如果是恰当方程请求解。
\begin{eqnarray*}
(4xy+y^3)dx + (6y+2x^2+3xy^2)dy=0.
\end{eqnarray*}

\vspace{0.2cm}

{\color{red}解答：原方程记为 $Pdx+Qdy=0$. 计算可得 
%\begin{eqnarray*}
$P'_y = 4x+3y^2, 
Q'_x = 4x+3y^2
$.  %\end{eqnarray*}
 根据恰当方程的判别法，由 $P'_y=Q'_x$ 可知这是一个恰当方程。设 $Pdx+Qdy=d\Phi$, 则
\begin{eqnarray}
&& \Phi'_x = P = 4xy+y^3, \label{eq3-1} \\
&& \Phi'_y = Q = 6y+2x^2+3xy^2. \label{eq3-2}
\end{eqnarray}
由等式 (\ref{eq3-1}) 对 $x$ 积分得 
\begin{eqnarray*}
\Phi = 2x^2y + xy^3+\varphi(y),  
\end{eqnarray*}
其中 $\varphi(y)$ 待定。代入等式 (\ref{eq3-2}) 得
\begin{eqnarray*}
\Phi'_y = 2x^2 + 3xy^2 + \varphi'(y) = 6y+2x^2+3xy^2. 
\end{eqnarray*}
于是 $\varphi'(y)=6y$, 可得 $\varphi(y)=3y^2+c$, 其中 $c$ 是任意常数。所以原方程的通解为
\begin{eqnarray*}
\Phi = 2x^2y + xy^3+3y^2+c=0.
\end{eqnarray*}

}

\vspace{0.2cm}

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\item  %第4题
考虑一阶线性微分方程 $y' + 2xy = x$. 
\begin{enumerate}
\item  使用积分因子 $m(x)=\exp(x^2)$ 求解。
\item  将原方程写成如下形式，然后使用分离变量法求解，
\begin{eqnarray*}
\frac{dy}{dx} = x(1-2y). 
\end{eqnarray*}
\end{enumerate}

\vspace{0.2cm}

{\color{red}解答：
\begin{enumerate}
\item  设 $m(x)=e^{(x^2)}$, 则 $m'(x)=2xm(x)$. 将 $m(x)$ 乘到方程两边，可得
\begin{eqnarray*}
e^{(x^2)}(y' + 2xy) = xe^{(x^2)}. 
\end{eqnarray*}
左边有原函数 $e^{(x^2)}y$, 右边有原函数 $\frac{1}{2}e^{(x^2)}$, 所以对上式两边积分，得
\begin{eqnarray*}
e^{(x^2)}y = \frac{1}{2}e^{(x^2)} +c, 
\end{eqnarray*}
其中 $c$ 是任意常数。所以通解为
\begin{eqnarray*}
y = \frac{1}{2} +ce^{-(x^2)}. 
\end{eqnarray*}

\item  分离变量可得
\begin{eqnarray*}
\frac{dy}{2y-1} = -xdx. 
\end{eqnarray*}
两边积分可得
\begin{eqnarray*}
\frac{1}{2}\ln |2y-1| = -\frac{1}{2}x^2 + c.  
\end{eqnarray*}
当 $y>\frac{1}{2}$ 时，化简可得 $y = \frac{1}{2} + ce^{-(x^2)}$, 其中 $c>0$.  

当 $y<\frac{1}{2}$ 时可得同样表达式，其中 $c<0$. 
当 $c=0$ 时恰好为特解 $y=\frac{1}{2}$. 

\end{enumerate}

}

\vspace{0.2cm}

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\item  %第5题
求解微分方程 $\frac{dy}{dx} = \frac{y-x}{y+3x}$.

\vspace{0.2cm}

{\color{red}解答：设变量代换 $y=xu$. 则原方程化为
\begin{eqnarray*}
x\frac{du}{dx} + u = \frac{xu-x}{xu+3x}.
\end{eqnarray*}
当 $x\neq 0$ 时，化简得到 
\begin{eqnarray*}
x\frac{du}{dx} = \frac{u-1}{u+3} -u = \frac{-u^2-2u-1}{u+3}.
\end{eqnarray*}
分离变量可得
\begin{eqnarray*}
\frac{u+3}{u^2+2u+1}du = \frac{-1}{x}dx.
\end{eqnarray*}
根据有理函数的积分方法，化成如下形式
\begin{eqnarray*}
\left[ \frac{1}{u+1} + \frac{2}{(u+1)^2} \right] du = \frac{-1}{x}dx.
\end{eqnarray*}
积分可得 
\begin{eqnarray*}
\ln |u+1| + \frac{-2}{u+1} = -\ln |x| + c, 
\end{eqnarray*}
其中 $c$ 是任意常数。代入 $u=y/x$ 可得
\begin{eqnarray*}
\ln \left\vert \frac{y}{x}+1\right\vert + \frac{-2x}{y+x} = -\ln |x| +c. 
\end{eqnarray*}
整理得到
\begin{eqnarray*}
\ln \left\vert y+x\right\vert - \frac{2x}{y+x} = c. 
\end{eqnarray*}

}

\vspace{0.2cm}

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\item  %第6题
考虑微分方程的初值问题 $\frac{dy}{dx} = xy+1, \, y(0)=1$.

\begin{enumerate}
\item  使用皮卡定理验证这个初值问题的解是存在且唯一的。
\item  写出皮卡序列的前三个函数。
\end{enumerate}

\vspace{0.2cm}

{\color{red}解答：
\begin{enumerate}
\item  这里 $f(x,y)=xy+1$, 这个二元函数在 $(x,y)$ 平面上是连续的。对任意正数 $a,b$, 在矩形区域 $-a<x<a, 1-b<y<1+b$ 上，函数 $f(x,y)=xy+1$ 满足李氏条件。因为 $f'_y=x$, 这只要取 $L=a$ 就可以了。

\item  先将这个初值问题写成积分方程，可得
\begin{eqnarray*}
y(x) = 1 + \int_0^x [ ty(t)+1 ] dt.
\end{eqnarray*}
于是皮卡序列的迭代公式和前三个函数分别是
\begin{eqnarray*}
y_{n+1}(x) &=& 1 + \int_0^x [ ty_n(t)+1 ] dt, \\
y_0(x) &=& 1, \\
y_1(x) &=& 1 + \int_0^x ( t\cdot 1+1 ) dt = 1+ x+\frac{x^2}{2}, \\
y_2(x) &=& 1 + \int_0^x [ t\cdot (1+t+\frac{t^2}{2})+1 ] dt =  1+x+\frac{x^2}{2} +\frac{x^3}{3}+\frac{x^4}{8}.
\end{eqnarray*}


\end{enumerate}

}

\vspace{0.2cm}

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\item  %第7题
求解微分方程 $\frac{dy}{dx} = y^2-1$. 并讨论在区域 $y>1$ 里的积分曲线的存在区间。
\vspace{0.2cm}

{\color{red}解答：首先看到 $y=\pm 1$ 是两个特解。当 $y\neq\pm 1$ 时，分离变量可得
\begin{eqnarray*}
\frac{dy}{y^2-1} = dx.
\end{eqnarray*}
根据有理函数的积分方法，先分解因式，再两边积分得到 
\begin{eqnarray*}
\frac{1}{2}\left(\frac{1}{y-1}-\frac{1}{y+1} \right) dy &=& dx, \\
\ln \left\vert \frac{y-1}{y+1} \right\vert &=& 2x+2c, 
\end{eqnarray*}
其中 $c$ 是任意常数。当 $y>1$ 时，由上式解得
\begin{eqnarray*}
y = \frac{1+e^{2x+2c}}{1-e^{2x+2c}}.
\end{eqnarray*}
因为 $y>1$, 所以 $x<-c$. 可见当 $x<-c$ 且 $x\to-c$ 时，$y\to+\infty$. 当 $x\to -\infty$ 时，$y>1$ 且 $y\to 1$. 这样可以画出在 $y>1$ 区域的积分曲线族。可见这些解的存在区间是 $(-\infty,-c)$. 

}

\vspace{0.2cm}

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\item  %第8题
考虑微分方程 $y=xp+p^2$, 其中 $p=\frac{dy}{dx}$. 
\begin{enumerate}
\item  使用微分法求解。
\item  使用 $p$-判别式求可能的奇解。
\item  按定义验证是不是奇解。
\end{enumerate}

\vspace{0.2cm}

{\color{red}解答：
\begin{enumerate}
\item  方程两边对 $x$ 求导，可得 $p=p+xp'+2pp'$. 化简得 $(x+2p)p'=0$. 于是 $p=-x/2$ 或 $p'=0$. 
\begin{enumerate}
\item  当 $p=-x/2$ 时，代入原方程，得到 $y=-x^2/2+x^2/4$, 即 $y=-x^2/4$. 验证可知是解。
\item  当 $p'=0$ 时可得 $p=c$ 是任意常数。代入原方程，得到通解 $y=cx+c^2$. 
\end{enumerate}

\item  将原方程写为 $F(x,y,p)=y-xp-p^2$, 按定义，$p$-判别式为如下联立方程组
\begin{eqnarray*}
F(x,y,p) &=& y-xp-p^2=0, \\
F'_p(x,y,p) &=& -x-2p=0.
\end{eqnarray*}
由第二式可得 $p=-x/2$, 代入第一式得到 $y+x^2/2-x^2/4=0$, 即 $y=-x^2/4$. 

\item  将通解和特解联立，
\begin{eqnarray*}
y &=& cx+c^2, \\
y &=& -x^2/4.
\end{eqnarray*}
求得交点的坐标为 $(x,y)=(-2c,-c^2)$. 求通解和特解在这一点的切线斜率，可得 
\begin{eqnarray*}
k_1 &=& c, \\
k_2 &=& -(-2c)/2=c. 
\end{eqnarray*}
因此通解和特解在这一点相切。所以 $y=-x^2/4$ 是奇解。 
 
\end{enumerate}

}

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\end{enumerate}


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